How do you solve $2 {tan}^{2} x + {sec}^{2} x = 2$ $2tan^2x+sec^2x=2$ ?

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How do you solve $2 {tan}^{2} x + {sec}^{2} x = 2$ $2tan^2x+sec^2x=2$ ?

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Answer 1 · 2016-09-18T04:06:56

This can be written as:

$\frac{2 {sin}^{2} x}{{cos}^{2} x} + \frac{1}{{cos}^{2} x} = 2$ $(2sin^2x)/cos^2x + 1/cos^2x = 2$ , using the identities $tan θ = \frac{sin θ}{cos θ}$ $tantheta = sin theta/costheta$ and $sec θ = \frac{1}{cos θ}$ $sectheta = 1/costheta$

Simplifying:

$\frac{2 {sin}^{2} x + 1}{{cos}^{2} x} = 2$ $(2sin^2x+ 1)/cos^2x = 2$

$2 {sin}^{2} x + 1 = 2 {cos}^{2} x$ $2sin^2x + 1 = 2cos^2x$

Apply the identity ${sin}^{2} x + {cos}^{2} x = 1 \to {sin}^{2} x = 1 - {cos}^{2} x$ $sin^2x + cos^2x = 1 -> sin^2x = 1 - cos^2x$ to the left-hand side.

$2 (1 - {cos}^{2} x) + 1 = 2 {cos}^{2} x$ $2(1 - cos^2x) + 1 = 2cos^2x$

$2 - 2 {cos}^{2} x + 1 = 2 {cos}^{2} x$ $2 - 2cos^2x + 1 = 2cos^2x$

$3 = 4 {cos}^{2} x$ $3 = 4cos^2x$

$\frac{3}{4} = {cos}^{2} x$ $3/4 = cos^2x$

$\pm \frac{\sqrt{3}}{2} = cos x$ $+-sqrt(3)/2 = cosx$

$x = \frac{π}{6}, \frac{5 π}{6}, \frac{7 π}{6} and \frac{11 π}{6}$ $x = pi/6, (5pi)/6, (7pi)/6 and (11pi)/6$

Note that these solutions are only those that are in the interval $0 \leq x < 2 π$ $0 ≤ x < 2pi$ .

Hopefully this helps!

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How do you solve $2 {tan}^{2} x + {sec}^{2} x = 2$ $2tan^2x+sec^2x=2$ ?

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